This can be solve using momentum balance
Momentum = mv
Where m is the mass of the object
V is the velocity of the object
(0.012 kg) ( 150 m/s) + ( 8.5 kg) ( 0 m/s) = ( 0.012 kg) ( -100 m/s) + (v) ( 8.5 kg)
V = 0.35 m/s
Answer: The velocity of concrete block is 0.353 m/s
Explanation:
To calculate the velocity of the concrete block after the collision, we use the equation of law of conservation of momentum, which is:
[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2[/tex]
where,
[tex]m_1[/tex] = mass of rubber bullet = 0.012 kg
[tex]u_1[/tex] = Initial velocity of rubber bullet = 150 m/s
[tex]v_1[/tex] = Final velocity of rubber bullet = -100 m/s
[tex]m_2[/tex] = mass of concrete block = 8.5 kg
[tex]u_2[/tex] = Initial velocity of concrete block = 0 m/s
[tex]v_2[/tex] = Final velocity of concrete block = ?
Putting values in above equation, we get:
[tex](0.012\times 150)+(8.5\times 0)=(0.012\times (-100))+(8.5\times v_2)\\\\v_2=\frac{1.8+1.2}{8.5}=0.353m/s[/tex]
Hence, the velocity of concrete block is 0.353 m/s
